Abstract
The harmonic index of a graph () is defined as the sum of the weights for all edges of , where is the degree of a vertex in . In this paper, we show that and , where is a quasi-tree graph of order and diameter . Indeed, we show that both lower bounds are tight and identify all quasi-tree graphs reaching these two lower bounds.
1. Introduction
Let be a simple connected graph with vertex set and edge set of order (). The harmonic index of , first appeared in [1], is defined as , where for , is the degree of in . For , the distance between and is shown by . Also, is the diameter of and .
The applications of the harmonic index in various chemical disciplines have been demonstrated in [2–4]. Also, several studies have focused on graph theoretical properties of the harmonic index, see, for example, [5–9]. For a broad overview, we refer to [10].
A connected graph is a quasi-tree graph if is not a tree and there exists a vertex such that is a tree. A graph is called unicyclic if it contains only one cycle. Obviously, every unicyclic graph is a quasi-tree graph. Many researchers have studied topological indices of quasi-tree graphs. See, for example, [11–16].
Liu [17] found a relation between harmonic index and diameter of a graph. He proved that if and is a connected graph of order , then and . Also, a lower bound was found for trees. If is a tree of order , then and . Thereby, Liu [17] proposed the following conjecture.
Conjecture 1. Let be a connected graph with order ; then,
Jerline and Michaelraj [18, 19] found a sharper bound for unicyclic graphs. They showed that if is a unicyclic graph of order , then and . They introduced a family of graphs, , which is a set of graphs obtained from by attaching one pendant vertex and a path of length to two diametrically nonadjacent vertices of (see Figure 1). Then, they proposed the following conjecture [18].
Conjecture 2. Let be a simple connected graph, which is not a tree, of order . Then,where equality holds if and only if .
They also show that the inequalities of the above conjecture are not true for a graph of order 6, namely, .
Suppose is a graph of order 4 which is obtained from by deleting an edge. Also, for , let be a family of graphs obtained from by attaching two paths of lengths and to two nonadjacent vertices of (see Figure 1). We will show that the inequality holds for all quasi-tree graphs except and the graph which is obtained by attaching two pendant vertices to two vertices of . Also, the equality holds for .
Two main theorems of this paper are as follows.
Theorem 1. Let be a quasi-tree graph of order . Then, . The equality holds if and only if or .
Theorem 2. Let be a quasi-tree graph of order and . Then, . The equality holds if and only if or .
In Section 2, we prove the lemmas that will be used in Section 3, where we prove the main theorems.
All graphs considered in this paper are finite, undirected, connected, and simple. Let be a graph and and a path of ; then, by and , we mean the graph obtained from by deleting the vertex and the vertices of , respectively. For all other notation and definitions not given here, the readers are referred to [20].
2. Preliminaries
Lemma 1. For , the two variables’ functionis positive.
Proof. Given , the terms inside parentheses are positive.
Lemma 2. for every .
Proof. The first inequality is valid since .
Let . Then, . So, is an increasing function and for every .
3. Proof of the Main Theorems
In this section, we will show that Conjecture 2 is true for all quasi-tree graphs. Also, in our proof, it will be shown that the equality in both inequalities hold whenever is the graph .
Lemma 3. Let be a quasi-tree graph of order , where , such that . Then,
Proof. If , then should be the complete graph, . In this case, and, by an easy calculation, and both inequalities hold.
If , then, since is not a quasi-tree graph, . Also, since is not a tree, and . Hence, . So, both inequalities hold for (see Figure 2).
Suppose and is the vertex that is a tree. Since is a tree of order 4, or 2. Also, since is a quasi-tree graph, it is not a complete graph, and hence, or 2.
If , then . If , then and . All quasi-tree graphs of order 5 and their harmonic indices are shown in Figure 3. As it is seen, all graphs hold both inequalities except when .
Suppose and is the vertex that is a tree. So, is one of , or , where is obtained by attaching a new pendant vertex to a pendant vertex of .
If , then . Since and is not , so , , and . On the contrary, for every edge of , is at least , where and are the degree of and in , respectively. So,The second and third inequalities hold by Lemma 2.
If , then the graph is one of the graphs shown in Figure 4 in which their harmonic indices are calculated. Also, , so and . So, for every graph both inequalities hold, except when . Also, the equality holds when .
If , then the graph is one of the graphs shown in Figure 5 in which their harmonic indices are calculated. Also, , so and . Obviously, for every graph, both inequalities hold.
Theorem 3. Let be a quasi-tree graph with vertices. Then, . The equality holds if and only if or .
Proof. By induction on , if , then Theorem 3 implies that the inequality is true, unless when . Also, by Theorem 3, the equality holds when .
Let be a quasi-tree graph with vertices. Suppose is a vertex of such that is a tree. Let be the diametrical path of . There are three cases as follows. Case 1. There exists such that . Since is not in diametrical path of and , and is a quasi-tree graph too. Suppose and . Then, by induction hypothesis, If , then is one of the graphs shown with their harmonic indices in Figure 6. In this case, and . Hence, the inequality holds. Case 2. Every vertex of is of degree at least 2, and there exists such that . Similar to the previous case, . Suppose . Note that , otherwise , and hence, , a contradiction. Since , it is possible that be a tree. There exist three subcases as follows: Subcase 2.1. is not a tree and are not adjacent in . By the hypothesis, there exist at most two vertices of degree 1 in which are and . If , then , and since are not adjacent, , a contradiction. So, has at most one neighbor of degree 1. Same argument is valid for , so If , then is one of the graphs which are shown with their harmonic indices in Figure 7. In this case, and . Subcase 2.2. is not a tree and are adjacent in . If , then Since , the last inequality is obtained from Lemma 1. If , then is the graph which is shown with its harmonic index in Figure 8. In Subcase 2.1, and the inequality holds. Subcase 2.3. is a tree. Since is also a tree, by counting the number of edges and vertices, it is obtained that . This means that is a unicyclic graph and as proved by Theorem 3 in[19], , in which equality holds if . Case 3. Every vertex of is of degree at least 2 and if and , then . Since is a tree, every pendant vertex of is in . So, is a path and . Also, since is not a tree . If , then is a unicyclic graph and as proposed by Theorem 3 in [19], , with equality holds if . So, there exist two subcases as follows. Subcase 3.1. In this case, is one of the graphs in Table 1. In all cases, . As it is shown, the equality holds when . Subcase 3.2 (). Suppose such that . If , then the diametrical path is longer than the path , a contradiction. So, , which is another contradiction. Hence, this case does not happen. The inequality holds in all cases and equality holds if and only if or .
Theorem 4. Let be a quasi-tree graph with vertices, then
The equality holds if and only if or .
Proof. The proof is similar to the proof of Theorem 3. By induction on , if , then Theorem 3 implies that the inequality holds unless when .
Let be a quasi-tree graph with vertices. Suppose is a vertex of such that is a tree. Let be the diametrical path of . Since is a tree, . There exist three cases as follows. Case 1. There exists such that . Since is not in diametrical path of , . Suppose . So, such as Case 1 in the proof of Theorem 3, if , then, by induction hypothesis, If , then is one of the graphs which is shown with their harmonic indices in Figure 6. In this case, and . Hence, the inequality holds. Case 2. Every vertex of is of degree at least 2, and there exists a vertex such that . From the previous case, . Suppose . Since , it is possible that be a tree. There exist three subcases as follows. Subcase 2.1. is not a tree and are not adjacent in . By the same argument as in Subcase 2.1 of Theorem 3, If , then , and is one of the graphs which are shown with their harmonic indices in Figure 7. Subcase 2.2. is not a tree and are adjacent in . By Subcase 2.2 of Theorem 3, So, by the induction hypothesis, If , then and is the graph which is shown with its harmonic index in Figure 8 and the inequality holds. Subcase 2.3. is a tree. By the same argument as in the proof of Subcase 2.3 of Theorem 3, is a unicyclic graph and by Theorem 3 in [18], , with equality when . Case 3. Every vertex of is of degree at least 2, and if and , then . By the same argument as in Case 3 of Theorem 3, is a path and . Also, since G is not a tree, . If , then G is a unicyclic graph and by Theorem 3 in[18], , with equality if or . Also, by Subcase 3.2 of Theorem 3, . So, and is one of the graphs in Table 1. Obviously, , for all of them. As in Theorem 3, the inequality holds in all cases and the equality will be satisfy if and only if or .
Data Availability
No data were used to support this study.
Disclosure
An earlier version of this manuscript has been presented as arXiv in Semantic Scholar according to this link.
Conflicts of Interest
The authors declare that they have no conflicts of interest.